Transfer function and stability from [1 0; 1 1]

For $\dot{x} = Ax + Bu$, $y = Cx + Du$:

$$\frac{Y(s)}{U(s)} = C(sI - A)^{-1}B + D = \frac{C\,\operatorname{adj}(sI-A)\,B}{|sI-A|} + D$$ $$A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix},\quad B = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\quad C = \begin{bmatrix} 1 & 1 \end{bmatrix}$$

Step 1: $(sI-A)$ and determinant

$$sI - A = \begin{bmatrix} s-1 & 0 \\ -1 & s-1 \end{bmatrix},\qquad |sI-A| = (s-1)^2 = s^2 - 2s + 1$$

Step 2: Inverse

$$(sI-A)^{-1} = \frac{1}{(s-1)^2}\begin{bmatrix} s-1 & 0 \\ 1 & s-1 \end{bmatrix}$$

Step 3: Transfer function

$$\begin{aligned} (sI-A)^{-1}B &= \frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ 1 + s - 1 \end{bmatrix} = \frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ s \end{bmatrix} \\ \frac{Y(s)}{U(s)} &= \begin{bmatrix} 1 & 1 \end{bmatrix}\frac{1}{(s-1)^2}\begin{bmatrix} s-1 \\ s \end{bmatrix} = \frac{2s-1}{(s-1)^2} = \frac{2s-1}{s^2-2s+1} \end{aligned}$$

Stability

Characteristic equation: $(s-1)^2 = 0$, so there is a repeated pole at $s = +1$ in the right half of the s-plane (also visible because $A$ is lower triangular with diagonal entries 1, 1).

The impulse response contains terms $e^{t}$ and $te^{t}$, which grow without bound. The system is unstable.

Answer: $\dfrac{Y(s)}{U(s)} = \dfrac{2s-1}{(s-1)^2}$; poles at $s = 1, 1$, zero at $s = 0.5$; unstable.